2.3.A,fA,fU071110

2.3.A,fA,fU071110 - 1 2.3. EBRT, IBRT in A,fA,fU. We redo...

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1 2.3. EBRT, IBRT in A,fA,fU. We redo section 2.2 for the signature A,fA,fU, with the same five BRT settings (SD,INF), (ELG SD,INF), (ELG,INF), (EVSD,INF), (MF,INF). After we treat these five BRT settings, we then treat the five corresponding unary BRT settings (SD[1],INF), (ELG[1] SD[1],INF), (ELG[1],INF), (EVSD[1],INF), (MF[1],INF). These are the same except that we restrict to the 1-ary functions only. There is quite a lot of difference between the unary settings and the multivariate settings; this was not the case in section 2.2, with just A,fA. We begin with EBRT in A,fA,fU. The 8 A,fA,fU pre elementary inclusions are as follows (see Definition 1.1.35). A fA fU = . A fA fU = U. A fA fU. fA A fU. fU A fA. A fA fU. A fU fA. fA fU A. The 6 A,fA,fU elementary inclusions are as follows (see Definition 1.1.36). A fA = . A fU = U. A fU. fU A fA. A fU fA. fA A. We will use Theorem 2.2.1, and the Complementation Theorem from section 1.3. In fact, we need the following strengthening of the Complementation Theorem. THEOREM 2.3.1. Let f SD and B INF. There exists A INF, A B, such that A fA = and B A fA. Moreover, this is provable in RCA 0 . Proof: Let f,B be as given. We inductively define A B as follows. Suppose the elements of A from 0,1,. ..,n-1 have been defined, n 0. We put n in A if and only if n B and
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2 n is not the value of f at arguments from A less than n. Then A is as required, using f SD. QED THEOREM 2.3.2. For all f EVSD there exists A INF such that A fA = , A fN = N. Moreover, this is provable in RCA 0 . Proof: Let f EVSD be k-ary. Let n be such that |x| n f(x) |x|. We define A inductively. First put [0,n]\fN in A. For m > n, put m in A if and only if m = f(x) for no |x| < m. Then [n+1, ) A fA. Also [0,n] A fA. Hence A fN = N. Suppose m A fA. If m n then by construction, m [0,n]\fN, contradicting m fA. Hence m > n. Let m = f(x), x A k . If |x| m then f(x) > |x| m, which is a contradiction. Hence |x| < m, and so m A by construction. This contradicts m A. QED THEROEM 2.3.3. Let k 2. There exists k-ary f ELG SD such that N\fN = {0}. There exists k-ary f ELG such that fN = N. Proof: For all n 1, let f n :[2 n ,2 n+1 ) k [2 n+1 ,2 n+2 ) be onto. Let f be the union of the f n extended as follows. For x not yet defined, set f(x) = 1 if |x| = 0; 2 if |x| = 1; 3 if |x| = 2; 2|x| if |x| 3. Then fN = N\{0} and f ELG SD. Let g be the union of the f n extended as follows. For x not yet defined, set f(x) = 0 if |x| = 0; 1 if |x| = 1; 2 if |x| = 2; 3 if |x| = 3; 2|x| if |x| 4. Then fN = N and f ELG. QED SETTINGS: (SD,INF), (ELG SD,INF), (ELG,INF), (EVSD,INF), (MF,INF).
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2.3.A,fA,fU071110 - 1 2.3. EBRT, IBRT in A,fA,fU. We redo...

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